Question

我是部分UI开发人员，现在我需要转向Hibernate开发。今天我在尝试使用UNION ALL时遇到了HQL问题，这里是HQL：

SELECT COUNT(DISTINCT users.userId) AS totalSize FROM (SELECT DISTINCT 
d1.sponsor.id AS userId FROM Dating d1 WHERE d1.invitee.id = ? UNION ALL 
SELECT DISTINCT d2.invitee.id AS userId FROM Dating d2 WHERE d2.sponsor.id = ?) 
AS users

它显示如下错误：

org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: ( near line 1, column 55

然后我尝试将此HQL转换为原始SQL：

SELECT COUNT(DISTINCT users.userId) AS totalSize FROM (SELECT DISTINCT 
d1.sponsorId AS userId FROM mmy_dating d1 WHERE d1.inviteeId = 6 UNION 
ALL SELECT DISTINCT d2.inviteeId AS userId FROM mmy_dating d2 WHERE 
d2.sponsorId = 6) AS users;

它通常显示正确的结果，所以我在想Hibernate是否不支持UNION ALL语法？

提前致谢。

Answer 1

此https://hibernate.atlassian.net/browse/HHH-1050的功能请求仍然有效。另请查看Hibernate Union alternatives

Answer 2

想分享，在我的情况下，我发现了一种避开这种情况的情况。这里唯一的规则是具有相同的类型，在这种情况下String，对应于返回列表，可以添加所需数量的表：

public List<String> findByCPForCNPJ(String query){
    TypedQuery<String> ccpf = manager.createQuery("select cpf from PessoaFisica where cpf like :pCpf", String.class);
    ccpf.setParameter("pCpf", "%" + query + "%");
    List<String> lista1 = ccpf.getResultList();

    TypedQuery<String> ccnpj = manager.createQuery("select cnpj from PessoaJuridica where cnpj like :pCnpj", String.class);
    ccnpj.setParameter("pCnpj", "%" + query + "%");

    lista1.addAll(ccnpj.getResultList());
    return lista1;
}

我希望我能为所有人做出贡献，祝你好运......

Hibernate是否支持UNION ALL？

2 个答案: